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Question

A variable plane passes through a fixed point (3,2,1) and meets x,y and z axes at A,B and C respectively. A plane is drawn parallel to yz−plane through A, a second plane is drawn parallel zx−plane through B and a third plane is drawn parallel to xy−plane through C. Then the locus of the point of intersection of these three planes, is :

A
x3+y2+z1=1
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B
x+y+z=6
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C
1x+1y+1z=116
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D
3x+2y+1z=1
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Solution

The correct option is D 3x+2y+1z=1
Let the equation of the plane be,
xa+yb+zc=1
So the points,
A=(a,0,0)B=(0,b,0)C=(0,0,c)
We know that the plane passes through (3,2,1)
So,
3a+2b+1c=1(1)
Now,
Plane parallel to yz plane and paasing through A,
x=a,
Similarly the other planes will be,
y=b,z=c
Point of intersection will be ,
(a,b,c)
From equation (1), the locus of point of intersection will be,
3x+2y+1z=1

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